[Math] limit of $x^2 \frac{\sin\frac{1}{x}}{\sin x}=0$ as $x\rightarrow 0$.

Question

Im having trouble trying to show that $\displaystyle\lim_{x \to 0} \frac{x^2\sin\frac{1}{x}}{\sin x}=0$. I tried l'hospitals rule since $\lim_{x \to 0} x^2 \sin(\frac{1}{x})=0$ and $\lim_{x \to 0} \sin x=0$. But when I work it out it doesn't work. I think im supposed to use the squeeze theorem. I know that $-1\leq \sin x\leq x$. But from here im stuck.

Answer 1

Hints:

i)

$$\lim x^2 \frac{\sin(1/x)}{\sin(x)} = \lim\frac{x}{\sin(x)}\cdot\lim x\sin(1/x). $$

ii)

$$ |x\sin(1/x)|\leq |x|. $$